Question

The decomposition of ${NH}_3$ on platinum surface is zero order. What are the rate of production of $N_2$ and $H_2$ in ${mole.lit}^{-1}.{sec}^{-1}$ if $K=2.5\times {10}^{-4}mole.li{t}^{-1}.{sec}^{-1}$?

• A. $3.75\times {10}^{-4},1.25\times {10}^{-4}$
• B. $1.25\times {10}^{-4},3.75\times {10}^{-4}$
• C. $2.5\times {10}^{-4},7.5\times {10}^{-4}$
• D. $1.25\times {10}^4,3.75\times {10}^{-4}$

Answer 1

Answer: (C)

$2.5\times {10}^{-4},7.5\times {10}^{-4}$

It is a zero-order reaction.

$2N{H}_3\to N_2+3H_2$

So,

$-\frac{1}{2}\frac{d\left[N{H}_3\right]}{dt}=+1\frac{d\left[N_2\right]}{dt}=+\frac{1}{3}\frac{d\left[H_2\right]}{dt}=2.5\times {10}^{-4}$ (Since its is a zero order reaction, Rate $=K$)

$\frac{d\left[N_2\right]}{dt}=2.5\times {10}^{-4}M/s$

$\frac{d\left[H_2\right]}{dt}=3\times 2.5\times {10}^{-4}=7.5\times {10}^{-4}M/s$